# `f''(t) = 3/(sqrt(t)), f(4) = 20, f'(4) = 7` Find `f`.

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### 1 Answer

`f''(t)=3/sqrt(t)`

`f'(t)=int(3/sqrt(t))dt`

`f'(t)=3(t^(-1/2+1)/(-1/2+1))+C_1`

`f'(t)=6sqrt(t)+C_1`

Now let's find constant C_1 given f'(4)=7,

`f'(4)=7=6sqrt(4)+C_1`

`7=12+C_1`

`C_1=-5`

`f'(t)=6sqrt(t)-5`

`f(t)=int(6sqrt(t)-5)dt`

`f(t)=6(t^(1/2+1)/(1/2+1))-5t+C_2`

`f(t)=6(t^(3/2)/(3/2))-5t+C_2`

`f(t)=4t^(3/2)-5t+C_2`

Now let's find constant C_2 given f(4)=20,

`f(4)=20=4(4^(3/2))-5(4)+C_2`

`20=4(2^2)^(3/2)-20+C_2`

`20=4(2^3)-20+C_2`

`20=32-20+C_2`

`C_2=8`

`:.f(t)=4t^(3/2)-5t+8`